Suppose we have a vector field $f(x, y) = (x \cos(y), y \cos(x))$ and a curve $C$ that is parameterized by $\alpha(t) = (2t, t)$ for $0 < t < \pi$. What is the line integral of $f$ along $C$ ? $ \int_C f \cdot d\alpha = $
Explanation: Given a vector field $f$, a parameterization $\alpha$, and bounds $t_0$ and $t_1$, we can calculate the line integral as follows: $ \int_C f \cdot d\alpha = \int_{t_1}^{t_2} f(\alpha(t)) \cdot \alpha'(t) \, dt$ Here, $f(x, y) = (x \cos(y), y \cos(x))$ and $\alpha(t) = (2t, t)$. $\begin{aligned} &f(\alpha(t)) = (2t \cos(t), t \cos(2t)) \\ \\ &\alpha'(t) = (2, 1) \end{aligned}$ Now we can rewrite our line integral as a single-variable integral. $ \int_C f \cdot d\alpha = \int_0^\pi (2t \cos(t), t \cos(2t)) \cdot (2, 1) \, dt$ Let's solve the integral. $\begin{aligned} \int_0^\pi (2t \cos(t), t \cos(2t)) \cdot (2, 1) \, dt &= \int_0^\pi 4t \cos(t) + t \cos(2t) \, dt \\ \\ &= \int_0^\pi t(4 \cos(t) + \cos(2t)) \, dt \end{aligned}$ We'll use integration by parts to evaluate this integral. [Details] Therefore: $\begin{aligned} &\int_0^\pi t(4 \cos(t) + \cos(2t)) \, dt \\ \\ &= \left[ 4t\sin(t) + \dfrac{t\sin(2t)}{2} \right]_0^\pi - \int_0^\pi 4\sin(t) + \dfrac{\sin(2t)}{2} \, dt \\ \\ &= 0 - \left[ -4\cos(t) - \dfrac{\cos(2t)}{4} \right]_0^\pi \\ \\ &= \left( -4 + \dfrac{1}{4} \right) - \left(4 + \dfrac{1}{4} \right) \\ \\ &= -8 \end{aligned}$ In conclusion, the line integral $ \int_C f \cdot d\alpha = -8$.